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3 years ago

Which interaction occurs when light goes into an object as heat energy?

Chemistry

2 answers:

xxTIMURxx [149]3 years ago

7 0

Answer:

Jdjejejehe sorry

Explanation:

DedPeter [7]3 years ago

3 0

Answer:

Absorption. Absorption occurs when photons from incident light hit atoms and molecules and cause them to vibrate. The more an object's molecules move and vibrate, the hotter it becomes. This heat is then emitted from the object as thermal energy.

Explanation:

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3 years ago

A steel beam that is 7.00 m long weighs 326 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

GrogVix [38]

Answer:

The answer to the question is

Suki is 1.12 m close to the end of the steel beam before the beam begins to tip

Explanation:

To solve the question we list out the known variables first

Length of steel beam, L = 7.00 m

Weight of steal beam W = 326 N

Distance between supports = 3.00 m

Weight of Suki = 555 N

By moments theory, the sum of moments about a point = 0

and assuming the steal beam is uniform, the weight of the beam will act at the center of the beam

As described the supports are 2.00 m from each end of the beam thus

taking moments about one of the support with Suki between the end of the beam and the support we have

Distance from center of beam to support = 1.5 m

Distance of Suki to the nearest support = x m

therefore 555 × x + 326 × 1.5 = 0

or x = -0.881 m on the other side of the support

Therefore Suki is (2 - 0.881) m or 1.12 m close to the end before the beam begins to tip

7 0

3 years ago

I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19

IRINA_888 [86]

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

For the 4.12 g sample

Moles of a substance is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}$

Number of molecules is given by

$nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}$

For the 19.37 g sample

$n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}$

Number of molecules is given by

$nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}$

$1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}$

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

7 0

3 years ago